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The line passing through the points $(1,2,-1)$ and $(3,-1,2)$ meets the yz-plane at which one of the following points?
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Verified Answer
The correct answer is:
$\left(0, \frac{7}{2},-\frac{5}{2}\right)$
Eqn. of line $\Rightarrow \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
i.e, $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=K($ say $)$
$\Rightarrow \mathrm{x}-1=2 \mathrm{~K} ; \mathrm{y}-2=-3 \mathrm{~K} ; \mathrm{z}+1=3 \mathrm{~K}$
$\Rightarrow \mathrm{x}=2 \mathrm{~K}+1 ; \mathrm{y}=-3 \mathrm{~K}+2 ; \mathrm{z}=3 \mathrm{~K}-1$
Since the line meets yz plane, $\mathrm{x}=0$
$2 \mathrm{~K}+1=0 \Rightarrow \mathrm{K}=\frac{-1}{2}$
$y=-3\left(\frac{-1}{2}\right)+2=\frac{3}{2}+2=\frac{7}{2}$
$z=3\left(\frac{-1}{2}\right)-1=\frac{-3}{2}-1=\frac{-5}{2}$
i.e, $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+1}{3}=K($ say $)$
$\Rightarrow \mathrm{x}-1=2 \mathrm{~K} ; \mathrm{y}-2=-3 \mathrm{~K} ; \mathrm{z}+1=3 \mathrm{~K}$
$\Rightarrow \mathrm{x}=2 \mathrm{~K}+1 ; \mathrm{y}=-3 \mathrm{~K}+2 ; \mathrm{z}=3 \mathrm{~K}-1$
Since the line meets yz plane, $\mathrm{x}=0$
$2 \mathrm{~K}+1=0 \Rightarrow \mathrm{K}=\frac{-1}{2}$
$y=-3\left(\frac{-1}{2}\right)+2=\frac{3}{2}+2=\frac{7}{2}$
$z=3\left(\frac{-1}{2}\right)-1=\frac{-3}{2}-1=\frac{-5}{2}$
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